Solve the following quadratic equations by factorization:

We have been given

$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$

$\frac{-11}{x^{2}-3 x-28}=\frac{11}{30}$

$-30=x^{2}-3 x-28$

$x^{2}-3 x+2=0$

$x^{2}-2 x-x+2=0$

$x(x-2)-1(x-2)=0$

$(x-1)(x-2)=0$

Therefore,

$x-1=0$

$x=1$

or,

$x-2=0$

$x=2$

Hence, $x=1$ or $x=2$.